The acceleration of gravity on or near the Earth's surface is 9.8 m/s² downward.
Is that right ? I don't hear any objection, so I'll assume that it is.
That means that during every second that gravity is the only force on an object,
the object either gains 9.8m/s of downward speed, or it loses 9.8m/s of upward
speed. (The same thing.)
If the rock starts out going up at 14.2 m/s, and loses 9.8 m/s of upward speed
every second, it runs out of upward gas in (14.2/9.8) = <em>1.449 seconds</em> (rounded)
At that point, since it has no more upward speed, it can't go any higher. Right ?
(crickets . . .)
Answer:
Final velocity = 7.677 m/s
KE before crash = 202300 J
KE after crash = 182,702.62 J
Explanation:
We are given;
m1 = 1400 kg
m2 = 4700 kg
u1 = 17 m/s
u2 = 0 m/s
Using formula for inelastic collision, we have;
m1•u1 + m2•u2 = (m1 + m2)v
Where v is final velocity after collision.
Plugging in the relevant values;
(1400 × 17) + (4700 × 0) = (1400 + 1700)v
23800 = 3100v
v = 23800/3100
v = 7.677 m/s
Kinetic energy before crash = ½ × 1400 × 17² = 202300 J
Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J
The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant = 8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg
Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec
A.) Electromagnetic Current
please mark me as the brainliest
